Đáp án:
Giải thích các bước giải:
$=\lim\dfrac{-2x+5}{\sqrt{x^2-2x+5}+x}$
$=\lim \dfrac{-2+\dfrac{5}{x}}{\sqrt{1-\dfrac{2}{x}+\dfrac{5}{x^2}}+1}$
$=\dfrac{\lim \left(-2+\dfrac{5}{x}\right)}{\lim\left(\sqrt{1-\dfrac{2}{x}+\dfrac{5}{x^2}}+1\right)}$
`=-1`
$\lim\left(\sqrt{x^2+2x+5}-x\right)$
$=\lim _{x\to \infty \:}\left(\dfrac{2+\dfrac{5}{x}}{\sqrt{1+\dfrac{2}{x}+\frac{5}{x^2}}+1}\right)$
`=2/2 = 1`
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$\lim _{x\to 1^-}\left(\dfrac{\sqrt{x+3}-2}{2x^2-3x+1}\right)$
$=\lim _{x\to \:1^-}\left(\dfrac{\dfrac{x-1}{\sqrt{x+3}+2}}{2x^2-3x+1}\right)$
$=\dfrac{1}{\left(\sqrt{1+3}+2\right)\left(2\cdot \:1-1\right)}$
`=1/4`
;
$\lim _{x\to 1}\left(\dfrac{-3}{x-4}\right)=1$
$\lim _{x\to 1^+}\left(\dfrac{-3}{x-4}\right)=1$
`->` Không liên tục
